Let three real numbers $$a, b, c$$ be in arithmetic progression and $$a+1, b, c+3$$ be in geometric progression. If $$a>10$$ and the arithmetic mean of $$a, b$$ and $$c$$ is 8, then the cube of the geometric mean of $$a, b$$ and $$c$$ is

Let a relation $$\mathrm{R}$$ on $$\mathrm{N} \times \mathbb{N}$$ be defined as: $$\left(x_1, y_1\right) \mathrm{R}\left(x_2, y_2\right)$$ if and only if $$x_1 \leq x_2$$ or $$y_1 \leq y_2$$. Consider the two statements:

(I) $$\mathrm{R}$$ is reflexive but not symmetric.

(II) $$\mathrm{R}$$ is transitive

Then which one of the following is true?

Let $$A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$$ and $$B=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} A)^{10}$$. Then, the sum of all the elements of the matrix $$B$$ is:

Let $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$$ and $$\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$$. If $$\vec{d}$$ is the unit vector in the direction of $$\vec{b}+\vec{c}$$ such that $$\vec{a} \cdot \vec{d}=1$$, then $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ is equal to